If I=∫12dx√(2x^3-9x^2+12x+4), then

Correct answer: (B) 1/9<I^2<1/8. fx=(1√(2x^3-9x^2+12x+4)f^(')x=-126x^2-18x+122x3-9x2+12x+4^32f^(')x=-6x-1x-222x3-9x2+12x+4^32f'x=0 ⇒x=1, 2f1=13 , f2=1√81)/3<fx<(1√8⇒∫1)/213dx<∫12fxdx<∫12(1√8dx⇒1)/3<I<(1√8 ⇒1)/9<I^2<1/8.

Q49 / 75 · JEE Main 2020 (08 Jan Shift 2) · Mathematics

If $I = \int_{1}^{2} \frac{d x}{\sqrt{2 x^{3} - 9 x^{2} + 12 x + 4}}$ , then

A
$\frac{1}{8} < I^{2} < \frac{1}{4}$
Correct answer
B
$\frac{1}{9} < I^{2} < \frac{1}{8}$
C
$\frac{1}{16} < I^{2} < \frac{1}{9}$
D
$\frac{1}{6} < I^{2} < \frac{1}{2}$

Solution

$f (x) = \frac{1}{\sqrt{2 x^{3} - 9 x^{2} + 12 x + 4}}$
$f^{'} (x) = \frac{- 1}{2} \frac{(6 x^{2} - 18 x + 12)}{{(2 x^{3} - 9 x^{2} + 12 x + 4)}^{\frac{3}{2}}}$

$f^{'} (x) = - \frac{6 (x - 1) (x - 2)}{2 {(2 x^{3} - 9 x^{2} + 12 x + 4)}^{\frac{3}{2}}}$

$f' (x) = 0 \Rightarrow x = 1, 2$
$f (1) = \frac{1}{3}$ , $f (2) = \frac{1}{\sqrt{8}}$

$\frac{1}{3} < f (x) < \frac{1}{\sqrt{8}}$

$\Rightarrow \int_{1}^{2} \frac{1}{3} d x < \int_{1}^{2} f (x) d x < \int_{1}^{2} \frac{1}{\sqrt{8}} d x$
$\Rightarrow \frac{1}{3} < I < \frac{1}{\sqrt{8}}$

$\Rightarrow \frac{1}{9} < I^{2} < \frac{1}{8}$ .

Hence, the correct answer is option: 'B'.

If I=∫12dx2x3-9x2+12x+4, then

Solution

More Definite Integration Questions

If $I = \int_{1}^{2} \frac{d x}{\sqrt{2 x^{3} - 9 x^{2} + 12 x + 4}}$ , then